Question

△ ABC is a right triangle right angled at A such that AB = AC and bisector of ∠ C intersects the side AB at D. Prove that AC + AD = BC.

Answer 1

Construct a triangle ABC where DE is perpendicular to BC

Consider the △ DAC and △ DEC

We know that

∠ BAC = ∠ DAC = 90^o

From the figure we know that CD bisects ∠ C

So we get

∠ DCA = ∠ DCE

We know that CD is common i.e. CD = CD

By AAS congruence criterion

△ DAC ≅ △ DEC

So we know that DA = DE …… (1)

AC = EC (c. p. c. t) ….. (2)

It is given that AB = AC

We know that the angles opposite to equal sides are equal

∠ B = ∠ C

We know that the sum of angles of △ ABC is 180^o.

∠ A + ∠ B + ∠ C = 180^o

By substituting the values

90^o + ∠ B + ∠ B = 180^o

On further calculation

2 ∠ B = 180^o – 90^o

2 ∠ B = 90^o

By division

∠ B = 45^o

Considering the △ BED

We know that ∠ BED = 90^o

So we can write it as

∠ BDE + ∠ B = 90^o

By substituting the values

∠ BDE + 45^o = 90^o

On further calculation

∠ BDE = 90^o – 45^o

By subtraction

∠ BDE = 45^o

It can be written as

∠ BDE = ∠ DBE = 45^o

We know that DE and BE are the equal sides of isosceles triangle

DE = BE …….. (3)

By comparing the equations (1) and (3)

We get

DA = DE = BE …… (4)

We know that BC = BE + EC

By considering the equations (ii) and (iv)

We get

BC = DA + AC

We can also write it as

AC + AD = BC

Therefore, it is proved that AC + AD = BC.