Question

in the adjoining figure seg ad perpendicular seg bc, b-d-c. show that ab^2-bd^2=ac^2-cd^2

Answer 1

Answer:

Given : $AD\perp BC$

To prove : $AB^2 - BD^2 = AC^2 - CD^2$

Proof:

By joining the points A with C and A with B ( construction)

We get two right triangles ADC and ADB,

Thus by the Pythagoras theorem,

$AB^2 = BD^2 + AD^2$  ---------(1)

And, $AC^2 = CD^2 + AD^2$  ---------(2)

Equation (1) - Equation (2),

We get,

$AB^2 - AC^2 = BD^2 - CD^2$

$AB^2 -BD^2 = AC^2 - CD^2$

Hence proved.

Answer 2

Step-by-step explanation:

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