Question

In the given figure, angle ACB=90° and CD is perpendicular to AB. Prove that the square of CD = BD x AD

Answer 1

We have to prove that AD^2=BD×DC.

In such problems convert what is to be proved into an equality of two ratios.

AD^2=BD×DC⇒ADBD=DCAD

You know that in similar triangles the ratios of the corresponding sides are equal.

So, try to find two triangles with AD and BD as one pair of corresponding sides and DC and AD as another pair of corresponding sides.

For this purpose, we need two triangles with one triangle having AD and DC as sides and another triangle having BD and AD as sides. The end points of the sides give us the triangles.

So, we select △ADCand△ADB.

It is easy to prove that in these two triangles ∠ADC=∠ADBand∠ACD=∠BAD.

Thus, we prove that the two angles are similar and therefore the ratios of their corresponding sides are equal.

Consequently, ADBD=DCAD

⇒AD^2=BD×DC.

Answer 2

Answer:

Step-by-step explanation:

To prove CD² = BD × AD

In Δ CAD, CA² = CD² + AD² .... (1)

Also in Δ CDB, CB² = CD² + BD² .... (2)

(1) + (2) we get

CA² + CB² = 2CD² + AD² + BD²

AB² = 2CD² + AD² + BD²

AB² - AD² = BD² + 2CD²

(AB + AD)(AB - AD) - BD² = 2CD²

(AB + AD)BD - BD² = 2CD²

BD(AB + AD - BD) = 2CD²

BD(AD + AD) = 2CD²

BD × 2AD = 2CD²

CD² = BD × AD

Hence proved.