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Triangles

Question
CBSEENMA9009818
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ΔABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB (see the given figure). Show that ∠BCD is a right angle.

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Solution

In ΔABC,

AB = AC (Given)

⇒ ∠ACB = ∠ABC (Angles opposite to equal sides of a triangle are also equal)

In ΔACD,

AC = AD

⇒ ∠ADC = ∠ACD (Angles opposite to equal sides of a triangle are also equal)

In ΔBCD,

∠ABC + ∠BCD + ∠ADC = 180º (Angle sum property of a triangle)

⇒ ∠ACB + ∠ACB +∠ACD + ∠ACD = 180º

⇒ 2(∠ACB + ∠ACD) = 180º

⇒ 2(∠BCD) = 180º

⇒ ∠BCD = 90º

Some More Questions From Triangles Chapter

In ΔABC, AD is the perpendicular bisector of BC (see the given figure). Show that ΔABC is an isosceles triangle in which AB = AC.

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ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively (see the given figure). Show that these altitudes are equal.

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ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see the given figure). Show that

(i) ΔABE ≅ ΔACF

(ii) AB = AC, i.e., ABC is an isosceles triangle.

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ΔABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB (see the given figure). Show that ∠BCD is a right angle.

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ABC is a right angled triangle in which ∠A = 90º and AB = AC. Find ∠B and ∠C.

Show that the angles of an equilateral triangle are 60º each.

AD is an altitude of an isosceles triangles ABC in which AB = AC. Show that

(i) AD bisects BC (ii) AD bisects ∠A.

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