Match the following if ABC is a right angled triangle at B and D is the foot of perpendicular from B to AC.

A D \times D C

No worries! We‘ve got your back. Try BYJU‘S free classes today!

A B^{2} + B C^{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

B C^{2} - A D \times D C

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Δ A D B

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

According to the theorem of geometric mean, $B D^{2} = A D \times D C$ .
According to pythagoras theorem, $A C^{2} = A B^{2} + B C^{2}$
By AA test, $Δ A B C \sim Δ A D B$
Applying pythagoras theorem in the triangle BCD, $B C^{2} = B D^{2} + C D^{2} \to C D^{2} = B C^{2} - B D^{2} \to C D^{2} = B C^{2} - A D \times D C$

Suggest Corrections

Similar questions

$△$ ABC is a right angled triangle, right angled at B. BD is perpendicular as shown. If AD: DC = 1:2, then $\frac{{A B}^{2}}{{B C}^{2}}$ is:

$△$ ABC is a triangle, right angled at B. BD is a perpendicular to AC as shown. Which of the following is true?

In $△ A B C$ , B is the right angle and BD is perpendicular to AC, then:

In the figure $△ A B C$ is a right angled triangle with right angle at B. BD is perpendicular to AC. Then which of the following options will hold true?

Join BYJU'S Learning Program

Match the following if ABC is a right angled triangle at B and D is the foot of perpendicular from B to AC.

According to the theorem of geometric mean, BD2=AD×DC. According to pythagoras theorem, AC2=AB2+BC2 By AA test, ΔABC∼ΔADB Applying pythagoras theorem in the triangle BCD, BC2=BD2+CD2→CD2=BC2−BD2→CD2=BC2−AD×DC