Internal angle bisector of A of triangle ABC, meets side BC at D. A line drawn through D perpendicular to AD intersects the side AC at P and the side AB at Q. If a, b, c represent the sides of ABC then. (a) AD=(2bc)/(b+c)cos A/2 (b) PQ=(4bc)/(b+c)sin A/2 (c) The triangle APQ is isosceles (d) AP is HM of b and c

Graham Beasley

Graham Beasley

Answered question

2022-07-16

Internal angle bisector of A of triangle Δ A B C, meets side BC at D. A line drawn through D perpendicular to AD intersects the side AC at P and the side AB at Q. If a, b, c represent the sides of Δ A B C then.
(a) A D = 2 b c b + c cos A 2
(b) P Q = 4 b c b + c sin A 2
(c) The triangle Δ A P Q is isosceles
(d) AP is HM of b and c
My approach is as follow

This is the rough image that I have drawn
cos A 2 A D = sin 90 o b 1
sin X b 2 = sin C P D
P D = A D tan A 2
sin X b 2 = sin C P D sin X b 2 = sin C A D tan A 2
The official answer is a,b,c and d.
I am not able to approach from here

Answer & Explanation

Ragazzonibw

Ragazzonibw

Beginner2022-07-17Added 15 answers

You can use sine-rule in A D C to find A D as in ( a ),
A D sin C = C D sin A / 2
where C D is found using B D : C D = c : b.
Next A D P A D Q by ASA congruence. Hence P D = Q D and A P = A Q ( c ).
Now in right A D P,
A P = A D sec A / 2 ( d )
Also
P Q = 2 P D = 2 A D tan A / 2 ( b )

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