Question

PQR is an acute angled triangle. PA bisects

P and meets QR in A. QP is produced
to E. PB is perpendicular to PA and meets QR produced in B. Through B, BE and BH
are drawn parallel to RP and PQ respectively, meeting QP and PR produced in E and
H respectively. Prove that PHBE is a rhombus.

Answer 1

Answer:

Given, Bisectors of ∠PQRand ∠PRS meet at point T.

To prove: ∠QTR=21∠QPR.

Proof,

∠TRS=∠TQR+∠QTR (Exterior angle of a triangle equals to the sum of the two interior angles.)

⇒∠QTR=∠TRS−∠TQR --- (i)

Also ∠SRP=∠QPR+∠PQR

2∠TRS=∠QPR+2∠TQR

∠QPR=2∠TRS−2∠TQR 

⇒21∠QPR=∠TRS−∠TQR --- (ii)

Equating (i) and (ii),

∴∠QTR=21∠QPR   [henceproved]