Question

in triangle abc ad perpendicular to BC and AD square is equal to BD into CD prove that a b square + AC square is equal to BD + CD square ​

Answer 1

Question:-

In ΔABC, AD is perpendicular to BC and AD² = BD × CD. Prove that AB² + AC² = (BD + CD)²

Solution:-

*Refer the attachment for figure.

It is given that, $\sf{AD\:\perp\:BC}$

In ΔADC

By Pythagoras Theorem

(Hypotenuse)² = (Perpendicular)² + (Base)²

OR

(H)² = (P)² + (B)²

=> (AC)² = (AD)² + (CD)² ____ (eq 1)

Similarly, In ΔADB

=> (AB)² = (AD)² + (BD)² _____ (eq 2)

Add (eq 1) & (eq 2)

=> (AC)² + (AB)² = (AD)² + (CD)² + (AD)² + (BD)²

=> (AC)² + (AB)² = 2(AD)² + (CD)² + (BD)² ____ (eq 3)

Also, given that AD² = BD × CD

Put value of AD² in (eq 3)

=> (AC)² + (AB)² = 2(BD × CD) + (CD)² + (BD)²

a² + b² + 2ab = (a + b)²

=> (AC)² + (AB)² = (BD + CD)²

Hence, proved

Answer 2

Answer:

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