Question

in the figure , D is a point on BC such that angle ABD = angle CAD
If AB =5 cm, AD = 4 cm and AC = 3 cm find
1) BC
2) A(∆ ACD) : A ( ∆ BCA)

Answer 1

I hope it will help you

Answer 2

Solution :-

In ∆ACD and ∆BCA we have,

→ ∠DCA = ∠ACB { common. }

→ ∠DAC = ∠ABC { given. }

so,

→ ∆ACD ~ ∆BCA { By AA similarity. }

then,

→ AC/BC = CD/CA = AD/BA { when two ∆'s are similar corresponding sides are in same proportion. }

putting given values now, we get,

→ AC/BC = AD/BA

→ 3/BC = 4/5

→ 4•BC = 3 * 5

→ BC = (15/4)

→ BC = 3.75 cm (i) (Ans.)

again,

→ CD/CA = AD/BA

→ CD/3 = 4/5

→ 5•CD = 3 * 4

→ CD = (12/5)

→ CD = DC = 2.4 cm (ii) (Ans.)

also, when two ∆'s are similar, the ratio of the areas is equal to the square of the ratio of their corresponding sides.

therefore,

→ A(∆ACD) : A(∆BCA) = AD² : BA² = 4² : 5² = 16 : 25 (iii) (Ans.)

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