Question

In the triangles ∆ ADC and ∆ CDB,

(1) AC = BC

(2) <ADC = <CDB (both 90°)

(3) CD is common side.

So, ∆ ADC is congruent to ∆ BDC.

$\bold{Prove\:\: that \:\: :}$
AD = CD​

Answer 1

In the triangles ∆ ADC and ∆ CDB,

(1) AC = BC

(2) <ADC = <CDB (both 90°)

(3) CD is common side.

So, ∆ ADC is congruent to ∆ BDC.

$\bold{Prove\:\: that \:\: :}$

AD = CD

∆ ADC is similar to ∆ BDC.

AC/BC = AD/CD = CD/BD

AC/BC = AD/CD

1 = AD/CD as AC = BC

AD = CD

Answer 2

Answer:

In the triangles Δ ADC and Δ CDB ,

∠ADC = ∠BDC [ 90° each ]

Given :

AC = BC

Hence ∠DAC = ∠DBC .

[ Base angles of an isosceles triangle ]

Δ ADC ≈ Δ BDC [ A.A criteria ]

Hence AD/CD = AC/BC

[ Ratio of corresponding sides are equal in similar triangles ]

Hence :-

AD/DC = AC/BC

Since it is given in the question that AC = BC ,

AC/BC = 1

Hence AD/DC = 1

⇒ AD = DC

Hence proved .

Step-by-step explanation:

The similarity of triangles was proved according to A.A criteria .

It states that when any two angles of two triangles are equal , then the triangles are said to be similar .

The similar triangles have the same ratio of corresponding sides .