In a right triangle $A B C$ , the perpendicular $B D$ on the hypotenuse $A C$ is drawn. Prove that
$(i) A C \times A D = A B^{2}$
$(i i) A C \times C D = B C^{2}$

Open in App

Solution

Given: In a right-angled triangle $△ A B C, B D ⊥ A C .$
To Prove: $A C \times A D = {A B}^{2}$
$A C \times C D = {B C}^{2}$

Proof: We draw a circle with $B C$ as diameter. Since $∠ B D C = 90^{\circ}$ the circle on $B C$ as diameter will pass through $D$ .

Again, $B C$ is a diameter and $A B ⊥ B C .$ $A B$ is tangent to the circle at $B$ . Since $A B$ is a tangent and $A D$ is a secant to the circle.
$\Rightarrow A C \times A D = A B^{2}$

Also,
$A C \times C D = A C \times (A C - A D) = {A C}^{2} - A C \times A D = {A C}^{2} - {A B}^{2} [By Using A C \times A D = {A B}^{2}] = {B C}^{2} [△ A B C is a right triangle]$

Hence proved, $A C \times C D = {B C}^{2} .$

Suggest Corrections

Similar questions

Q. In triangle ABC, AB=AC and BD is perpendicular to AC prove that

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

Q. Triangle

A B C

is right angled at

C

and

C D

is perpendicular to

A B

. Prove that

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

In a ΔABC, obtuse angled at B, if AD is perpendicular to CB produced, prove that AC = ${A B}^{2}$ + ${B C}^{2}$ + 2BC $\times$ BD

$△$ ABC is a right angled triangle, right angled at B. BD is a perpendicular as shown. Prove that:

${A B}^{2}$ = ${A D}^{2}$ + ${B C}^{2}$ - ${C D}^{2}$

Q. In triangle

A B C, A B = A C

and

B D

is perpendicular to

A C

. Prove that:

B D^{2} - C D^{2} = 2 (C D \times A D)

Join BYJU'S Learning Program

In a right triangle ABC, the perpendicular BD on the hypotenuse AC is drawn. Prove that(i)AC×AD=AB2(ii)AC×CD=BC2

In a right triangle $A B C$ , the perpendicular $B D$ on the hypotenuse $A C$ is drawn. Prove that
$(i) A C \times A D = A B^{2}$
$(i i) A C \times C D = B C^{2}$