Question

In triangle ABC, AD is perpendicular to BC, AD^2=BD×CD. Prove that triangle is a right triangle.

Answer 1

Given : In triangle ABC , AD is perpendicular to BC and AD² = BD.DC

To prove : BAC = 90°

Proof : in right triangles ∆ADB and ∆ADC
So, Pythagoras theorem should be apply ,
Then we have ,
AB² = AD² + BD² ----------(1)
AC²= AD²+ DC² ---------(2)

AB² + AC² = 2AD² + BD²+ DC²
= 2BD . CD + BD² + CD² [ ∵ given AD² = BD.CD ]
= (BD + CD )² = BC²
Thus in triangle ABC we have , AB² + AC²= BC²

hence triangle ABC is a right triangle right angled at A

∠ BAC = 90°

Answer 2

AD^2 = BD × CD

AD  ÷ CD  =  BD  ÷ AD

ΔADC ∼ ΔBDA (by SAS ∠D = 90°)

∠BAD = ∠ACD ;

∠DAC = ∠DBA (Corresponding angles of similar triangles)

∠BAD+ ∠ACD + ∠DAC + ∠DBA = 180°

⇒ 2∠BAD + 2∠DAC = 180°

⇒ ∠BAD + ∠DAC = 90°

∴ ∠A = 90°