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RS Aggarwal Solutions Class 9 Mathematics Solutions for Congruence Of Triangles And Inequalities In A Triangle Exercise 9A in Chapter 9 - Congruence Of Triangles And Inequalities In A Triangle
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△ 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.
Construct a triangle ABC where DE is perpendicular to BC
Consider the △ DAC and △ DEC
We know that
∠ BAC = ∠ DAC = 90o
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
∠ A + ∠ B + ∠ C = 180
By substituting the values
90 + ∠ B + ∠ B = 180
On further calculation
2 ∠ B = 180 - 90
2 ∠ B = 90
By division
∠ B = 45
Considering the △ BED
We know that ∠ BED = 90
So we can write it as
∠ BDE + ∠ B = 90
By substituting the values
∠ BDE + 45o = 90
On further calculation
∠ BDE = 90 - 45
By subtraction
∠ BDE = 45
It can be written as
∠ BDE = ∠ DBE = 45
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.
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