Distance from meeting the bisectors to the side of the quadrilateral












1












$begingroup$


In a quadrilateral ABCD, AC and BD are bisectors of < BAD and < ADC. If AC intercepts BD at P, AB = 6, CD = 3 and < APD = 135°, calculate the distance P to AD.



I even designed it but I did not find the solution;



enter image description here






geometry







share|cite|improve this question









$endgroup$












  • $begingroup$
    Your drawing is not accurate.
    $endgroup$
    – Moti
    Dec 19 '18 at 1:02












  • $begingroup$
    Bisectors are dividing the sides with a certain ratio - use this to get 3 equations for the sides and solve.
    $endgroup$
    – Moti
    Dec 19 '18 at 1:31










  • $begingroup$
    @Moti It is hard to draw an accurate figure if you don't know the answer to the problem.
    $endgroup$
    – Quang Hoang
    Dec 19 '18 at 2:56
















1












$begingroup$


In a quadrilateral ABCD, AC and BD are bisectors of < BAD and < ADC. If AC intercepts BD at P, AB = 6, CD = 3 and < APD = 135°, calculate the distance P to AD.



I even designed it but I did not find the solution;



enter image description here






geometry







share|cite|improve this question









$endgroup$












  • $begingroup$
    Your drawing is not accurate.
    $endgroup$
    – Moti
    Dec 19 '18 at 1:02












  • $begingroup$
    Bisectors are dividing the sides with a certain ratio - use this to get 3 equations for the sides and solve.
    $endgroup$
    – Moti
    Dec 19 '18 at 1:31










  • $begingroup$
    @Moti It is hard to draw an accurate figure if you don't know the answer to the problem.
    $endgroup$
    – Quang Hoang
    Dec 19 '18 at 2:56














1












1








1





$begingroup$


In a quadrilateral ABCD, AC and BD are bisectors of < BAD and < ADC. If AC intercepts BD at P, AB = 6, CD = 3 and < APD = 135°, calculate the distance P to AD.



I even designed it but I did not find the solution;



enter image description here






geometry







share|cite|improve this question









$endgroup$




In a quadrilateral ABCD, AC and BD are bisectors of < BAD and < ADC. If AC intercepts BD at P, AB = 6, CD = 3 and < APD = 135°, calculate the distance P to AD.



I even designed it but I did not find the solution;



enter image description here






geometry




geometry






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 18 '18 at 17:23









peta arantespeta arantes

647




647












  • $begingroup$
    Your drawing is not accurate.
    $endgroup$
    – Moti
    Dec 19 '18 at 1:02












  • $begingroup$
    Bisectors are dividing the sides with a certain ratio - use this to get 3 equations for the sides and solve.
    $endgroup$
    – Moti
    Dec 19 '18 at 1:31










  • $begingroup$
    @Moti It is hard to draw an accurate figure if you don't know the answer to the problem.
    $endgroup$
    – Quang Hoang
    Dec 19 '18 at 2:56


















  • $begingroup$
    Your drawing is not accurate.
    $endgroup$
    – Moti
    Dec 19 '18 at 1:02












  • $begingroup$
    Bisectors are dividing the sides with a certain ratio - use this to get 3 equations for the sides and solve.
    $endgroup$
    – Moti
    Dec 19 '18 at 1:31










  • $begingroup$
    @Moti It is hard to draw an accurate figure if you don't know the answer to the problem.
    $endgroup$
    – Quang Hoang
    Dec 19 '18 at 2:56
















$begingroup$
Your drawing is not accurate.
$endgroup$
– Moti
Dec 19 '18 at 1:02






$begingroup$
Your drawing is not accurate.
$endgroup$
– Moti
Dec 19 '18 at 1:02














$begingroup$
Bisectors are dividing the sides with a certain ratio - use this to get 3 equations for the sides and solve.
$endgroup$
– Moti
Dec 19 '18 at 1:31




$begingroup$
Bisectors are dividing the sides with a certain ratio - use this to get 3 equations for the sides and solve.
$endgroup$
– Moti
Dec 19 '18 at 1:31












$begingroup$
@Moti It is hard to draw an accurate figure if you don't know the answer to the problem.
$endgroup$
– Quang Hoang
Dec 19 '18 at 2:56




$begingroup$
@Moti It is hard to draw an accurate figure if you don't know the answer to the problem.
$endgroup$
– Quang Hoang
Dec 19 '18 at 2:56










1 Answer
1






active

oldest

votes


















2












$begingroup$

enter image description here



Let $BE = b, EA = a, BA=c$ and the distance from $P$ to the three edges of $triangle ABE$ be $r$. Then
$$r = frac{b+a-c}{2}tag{1}.$$
Since $BC$ is the angle bisector of $angle EBA$,
$$frac{a-6}{b} = frac{6}c,text{ or } (a-6)c = 6btag{2}.$$
Similarly
$$frac{b-3}{b} = frac3c,text{ or }(b-3)c = 3atag{3}.$$
Take $2(3)-(2)$ we have
$$c(2b-a) = 6(a-b)tag{4}.$$



On the other hand,
$$color{red}{frac{r}{a-6}} + color{blue}{frac{r}{b}} = color{red}{frac{BP}{BC}} + color{blue}{frac{PC}{BC}} = 1$$
so $$r(a+b-6) = (a-6)b.tag{5}$$
And similarly
$$r(a+b-3) = (b-3)a.tag{6}$$
Taking $(6)-(5)$ we obtain
$$ r = 2b - a.tag{7}$$
Replace $r$ from $(7)$ into $(1)$ we get
$$a+b-c = 2(2b-a),text{ or } c = 3(a-b).tag{8}$$
Now replace $(7)$ and $(8)$ into $(4)$ we get $r=2$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks Quag.. Your letter are changed. CD = 3 and AB = 6, so instead of B should be C, instead of A should be B and instead of C should be A. I do not know the inradius formula r = b + a-c / 2. Another question: how did you get to r / (a-6) + r / b = (BP / BC) + (PC + BC)? Thanks for the clarifications
    $endgroup$
    – peta arantes
    Dec 19 '18 at 13:05












  • $begingroup$
    That was the inradius for a right triangle. Notice the square at the lower left corner. for the other question, in $triangle BEC$ (in my picture), the horizontal $r$ starting from $P$ is parallel to $EC$, and we have $r/(a-6) = r/EC = BP/BC$. Similarly for the vertical $r$, and $r/b = PC/EBC$.
    $endgroup$
    – Quang Hoang
    Dec 19 '18 at 13:15










  • $begingroup$
    Understood. Thank you very much
    $endgroup$
    – peta arantes
    Dec 19 '18 at 14:06












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1 Answer
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1 Answer
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active

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active

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2












$begingroup$

enter image description here



Let $BE = b, EA = a, BA=c$ and the distance from $P$ to the three edges of $triangle ABE$ be $r$. Then
$$r = frac{b+a-c}{2}tag{1}.$$
Since $BC$ is the angle bisector of $angle EBA$,
$$frac{a-6}{b} = frac{6}c,text{ or } (a-6)c = 6btag{2}.$$
Similarly
$$frac{b-3}{b} = frac3c,text{ or }(b-3)c = 3atag{3}.$$
Take $2(3)-(2)$ we have
$$c(2b-a) = 6(a-b)tag{4}.$$



On the other hand,
$$color{red}{frac{r}{a-6}} + color{blue}{frac{r}{b}} = color{red}{frac{BP}{BC}} + color{blue}{frac{PC}{BC}} = 1$$
so $$r(a+b-6) = (a-6)b.tag{5}$$
And similarly
$$r(a+b-3) = (b-3)a.tag{6}$$
Taking $(6)-(5)$ we obtain
$$ r = 2b - a.tag{7}$$
Replace $r$ from $(7)$ into $(1)$ we get
$$a+b-c = 2(2b-a),text{ or } c = 3(a-b).tag{8}$$
Now replace $(7)$ and $(8)$ into $(4)$ we get $r=2$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks Quag.. Your letter are changed. CD = 3 and AB = 6, so instead of B should be C, instead of A should be B and instead of C should be A. I do not know the inradius formula r = b + a-c / 2. Another question: how did you get to r / (a-6) + r / b = (BP / BC) + (PC + BC)? Thanks for the clarifications
    $endgroup$
    – peta arantes
    Dec 19 '18 at 13:05












  • $begingroup$
    That was the inradius for a right triangle. Notice the square at the lower left corner. for the other question, in $triangle BEC$ (in my picture), the horizontal $r$ starting from $P$ is parallel to $EC$, and we have $r/(a-6) = r/EC = BP/BC$. Similarly for the vertical $r$, and $r/b = PC/EBC$.
    $endgroup$
    – Quang Hoang
    Dec 19 '18 at 13:15










  • $begingroup$
    Understood. Thank you very much
    $endgroup$
    – peta arantes
    Dec 19 '18 at 14:06
















2












$begingroup$

enter image description here



Let $BE = b, EA = a, BA=c$ and the distance from $P$ to the three edges of $triangle ABE$ be $r$. Then
$$r = frac{b+a-c}{2}tag{1}.$$
Since $BC$ is the angle bisector of $angle EBA$,
$$frac{a-6}{b} = frac{6}c,text{ or } (a-6)c = 6btag{2}.$$
Similarly
$$frac{b-3}{b} = frac3c,text{ or }(b-3)c = 3atag{3}.$$
Take $2(3)-(2)$ we have
$$c(2b-a) = 6(a-b)tag{4}.$$



On the other hand,
$$color{red}{frac{r}{a-6}} + color{blue}{frac{r}{b}} = color{red}{frac{BP}{BC}} + color{blue}{frac{PC}{BC}} = 1$$
so $$r(a+b-6) = (a-6)b.tag{5}$$
And similarly
$$r(a+b-3) = (b-3)a.tag{6}$$
Taking $(6)-(5)$ we obtain
$$ r = 2b - a.tag{7}$$
Replace $r$ from $(7)$ into $(1)$ we get
$$a+b-c = 2(2b-a),text{ or } c = 3(a-b).tag{8}$$
Now replace $(7)$ and $(8)$ into $(4)$ we get $r=2$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks Quag.. Your letter are changed. CD = 3 and AB = 6, so instead of B should be C, instead of A should be B and instead of C should be A. I do not know the inradius formula r = b + a-c / 2. Another question: how did you get to r / (a-6) + r / b = (BP / BC) + (PC + BC)? Thanks for the clarifications
    $endgroup$
    – peta arantes
    Dec 19 '18 at 13:05












  • $begingroup$
    That was the inradius for a right triangle. Notice the square at the lower left corner. for the other question, in $triangle BEC$ (in my picture), the horizontal $r$ starting from $P$ is parallel to $EC$, and we have $r/(a-6) = r/EC = BP/BC$. Similarly for the vertical $r$, and $r/b = PC/EBC$.
    $endgroup$
    – Quang Hoang
    Dec 19 '18 at 13:15










  • $begingroup$
    Understood. Thank you very much
    $endgroup$
    – peta arantes
    Dec 19 '18 at 14:06














2












2








2





$begingroup$

enter image description here



Let $BE = b, EA = a, BA=c$ and the distance from $P$ to the three edges of $triangle ABE$ be $r$. Then
$$r = frac{b+a-c}{2}tag{1}.$$
Since $BC$ is the angle bisector of $angle EBA$,
$$frac{a-6}{b} = frac{6}c,text{ or } (a-6)c = 6btag{2}.$$
Similarly
$$frac{b-3}{b} = frac3c,text{ or }(b-3)c = 3atag{3}.$$
Take $2(3)-(2)$ we have
$$c(2b-a) = 6(a-b)tag{4}.$$



On the other hand,
$$color{red}{frac{r}{a-6}} + color{blue}{frac{r}{b}} = color{red}{frac{BP}{BC}} + color{blue}{frac{PC}{BC}} = 1$$
so $$r(a+b-6) = (a-6)b.tag{5}$$
And similarly
$$r(a+b-3) = (b-3)a.tag{6}$$
Taking $(6)-(5)$ we obtain
$$ r = 2b - a.tag{7}$$
Replace $r$ from $(7)$ into $(1)$ we get
$$a+b-c = 2(2b-a),text{ or } c = 3(a-b).tag{8}$$
Now replace $(7)$ and $(8)$ into $(4)$ we get $r=2$.






share|cite|improve this answer









$endgroup$



enter image description here



Let $BE = b, EA = a, BA=c$ and the distance from $P$ to the three edges of $triangle ABE$ be $r$. Then
$$r = frac{b+a-c}{2}tag{1}.$$
Since $BC$ is the angle bisector of $angle EBA$,
$$frac{a-6}{b} = frac{6}c,text{ or } (a-6)c = 6btag{2}.$$
Similarly
$$frac{b-3}{b} = frac3c,text{ or }(b-3)c = 3atag{3}.$$
Take $2(3)-(2)$ we have
$$c(2b-a) = 6(a-b)tag{4}.$$



On the other hand,
$$color{red}{frac{r}{a-6}} + color{blue}{frac{r}{b}} = color{red}{frac{BP}{BC}} + color{blue}{frac{PC}{BC}} = 1$$
so $$r(a+b-6) = (a-6)b.tag{5}$$
And similarly
$$r(a+b-3) = (b-3)a.tag{6}$$
Taking $(6)-(5)$ we obtain
$$ r = 2b - a.tag{7}$$
Replace $r$ from $(7)$ into $(1)$ we get
$$a+b-c = 2(2b-a),text{ or } c = 3(a-b).tag{8}$$
Now replace $(7)$ and $(8)$ into $(4)$ we get $r=2$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 19 '18 at 2:55









Quang HoangQuang Hoang

13.2k1233




13.2k1233












  • $begingroup$
    Thanks Quag.. Your letter are changed. CD = 3 and AB = 6, so instead of B should be C, instead of A should be B and instead of C should be A. I do not know the inradius formula r = b + a-c / 2. Another question: how did you get to r / (a-6) + r / b = (BP / BC) + (PC + BC)? Thanks for the clarifications
    $endgroup$
    – peta arantes
    Dec 19 '18 at 13:05












  • $begingroup$
    That was the inradius for a right triangle. Notice the square at the lower left corner. for the other question, in $triangle BEC$ (in my picture), the horizontal $r$ starting from $P$ is parallel to $EC$, and we have $r/(a-6) = r/EC = BP/BC$. Similarly for the vertical $r$, and $r/b = PC/EBC$.
    $endgroup$
    – Quang Hoang
    Dec 19 '18 at 13:15










  • $begingroup$
    Understood. Thank you very much
    $endgroup$
    – peta arantes
    Dec 19 '18 at 14:06


















  • $begingroup$
    Thanks Quag.. Your letter are changed. CD = 3 and AB = 6, so instead of B should be C, instead of A should be B and instead of C should be A. I do not know the inradius formula r = b + a-c / 2. Another question: how did you get to r / (a-6) + r / b = (BP / BC) + (PC + BC)? Thanks for the clarifications
    $endgroup$
    – peta arantes
    Dec 19 '18 at 13:05












  • $begingroup$
    That was the inradius for a right triangle. Notice the square at the lower left corner. for the other question, in $triangle BEC$ (in my picture), the horizontal $r$ starting from $P$ is parallel to $EC$, and we have $r/(a-6) = r/EC = BP/BC$. Similarly for the vertical $r$, and $r/b = PC/EBC$.
    $endgroup$
    – Quang Hoang
    Dec 19 '18 at 13:15










  • $begingroup$
    Understood. Thank you very much
    $endgroup$
    – peta arantes
    Dec 19 '18 at 14:06
















$begingroup$
Thanks Quag.. Your letter are changed. CD = 3 and AB = 6, so instead of B should be C, instead of A should be B and instead of C should be A. I do not know the inradius formula r = b + a-c / 2. Another question: how did you get to r / (a-6) + r / b = (BP / BC) + (PC + BC)? Thanks for the clarifications
$endgroup$
– peta arantes
Dec 19 '18 at 13:05






$begingroup$
Thanks Quag.. Your letter are changed. CD = 3 and AB = 6, so instead of B should be C, instead of A should be B and instead of C should be A. I do not know the inradius formula r = b + a-c / 2. Another question: how did you get to r / (a-6) + r / b = (BP / BC) + (PC + BC)? Thanks for the clarifications
$endgroup$
– peta arantes
Dec 19 '18 at 13:05














$begingroup$
That was the inradius for a right triangle. Notice the square at the lower left corner. for the other question, in $triangle BEC$ (in my picture), the horizontal $r$ starting from $P$ is parallel to $EC$, and we have $r/(a-6) = r/EC = BP/BC$. Similarly for the vertical $r$, and $r/b = PC/EBC$.
$endgroup$
– Quang Hoang
Dec 19 '18 at 13:15




$begingroup$
That was the inradius for a right triangle. Notice the square at the lower left corner. for the other question, in $triangle BEC$ (in my picture), the horizontal $r$ starting from $P$ is parallel to $EC$, and we have $r/(a-6) = r/EC = BP/BC$. Similarly for the vertical $r$, and $r/b = PC/EBC$.
$endgroup$
– Quang Hoang
Dec 19 '18 at 13:15












$begingroup$
Understood. Thank you very much
$endgroup$
– peta arantes
Dec 19 '18 at 14:06




$begingroup$
Understood. Thank you very much
$endgroup$
– peta arantes
Dec 19 '18 at 14:06


















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