- ACB-90o and CD AB- prove that CD2-BD - AD-

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Similar Triangles

∠ ACB=90o and...

Question

$∠ A C B = 90^{o}$ and $C D ⊥ A B$ , prove that $C D^{2} = B D \times A D$ .

Open in App

Solution

Suggest Corrections

Similar questions

In ΔABC, AB = AC and BD ⊥ AC prove that BD² + CD² = 2AC.CD.

∠ A C B = 90^{0}

C D ⊥ A B

\frac{{C B}^{2}}{{C A}^{2}} = \frac{B D}{A D}

△ A B C, ∠ A C B = 90^{o}

C D ⊥

A B

D E ⊥

C B

C D^{2} \times A C = A D \times A B \times D E

In the given figure, $∠ A C B = 90^{o} a n d C D ⊥ A B$ .

Prove that $\frac{B C^{2}}{A C^{2}} = \frac{B D}{A D}$ .

A B C D

∠ A D C = 90^{o}

A B = 9 c m

B C = A D = 6 c m

C D = 3 c m

∠ A C B = 90^{o}

Join BYJU'S Learning Program