Mathematics
In triangle ABC; angle ABC = 90° and P is a point on AC such that ∠PBC = ∠PCB. Show that : PA = PB.
Answer
In the right angled triangle ABC,
Let, ∠ACB = x
From figure,
⇒ ∠PCB = ∠ACB = x
⇒ ∠PBC = ∠PCB = x
By angle sum property of triangle,
⇒ ∠ABC + ∠ACB + ∠BAC = 180°
⇒ 90° + x + ∠BAC = 180°
⇒ ∠BAC = 180° - 90° - x = 90° - x
⇒ ∠BAP = ∠BAC = 90° - x.
From figure,
⇒ ∠ABP = ∠ABC - ∠PBC = 90° - x.
∴ ∠BAP = ∠ABP = 90° - x …..(1)
In △ BAP,
⇒ ∠BAP = ∠ABP [From (1)]
∴ PB = PA (Sides opposite to equal angles are equal)
Hence, proved that PA = PB.
Related Questions
In quadrilateral ABCD, side AB is the longest and side DC is the shortest. Prove that :
(i) ∠C > ∠A
(ii) ∠D > ∠B
In triangle ABC, side AC is greater than side AB. If the internal bisector of angle A meets the opposite side at point D, prove that : ∠ADC is greater than ∠ADB.
In isosceles triangle ABC, sides AB and AC are equal. If point D lies in base BC and point E lies on BC produced (BC being produced through vertex C), prove that :
(i) AC > AD
(ii) AE > AC
(iii) AE > AD
Given : ED = EC
Prove : AB + AD > BC.
