Careers360 Logo
ask-icon
share
    Top Private MBA Colleges in Rajasthan 2026: Fees, Placements, Ranking, Admission

    CAT Important Theorem of Triangles (Part 2) - Practice Questions & MCQ

    Edited By admin | Updated on Oct 04, 2023 04:20 PM | #CAT

    Quick Facts

    • 5 Questions around this concept.

    Solve by difficulty

    Two sides AB and BC, and the median AD of $\triangle$ABC are correspondingly equal to the two sides PQ and QR, and the median PM of $\triangle$PQR. Then:

    In the given figure,

    ABCD is a square, M is the midpoint of AB, and PQ $\perp$ CM meets AD at P and BC produced at Q. Then:

    In the given figure,

    ABC is a triangle, right angled at B. If BCDE is a square on side BC and ACFG is a square on AC. Then:

    Let P be a point equidistant from two intersecting lines L and M at point A. Then:

    Concepts Covered - 1

    Important Theorem of Triangles (Part 2)

    Theorem 1: The angles opposite to two equal sides of a triangle are equal.

    Given: Let a \triangle A B C\text{ in which }A B=A C

    To prove: \angle B=\angle C

    Construction: Draw AD, the bisector of \angle A, to meet BC in D.

    Proof: 

    {\color{Blue} \text{In } \triangle A B D\text{ and }\triangle A C D,\text{ we have,}}

    {\color{Blue} A B=A C \quad(\text { given })}

    {\color{Blue} AD=AD \quad(\text { common })}

    {\color{Blue} \angle B A D=\angle C A D \quad(\text { by construction })}

    {\color{Blue} \therefore \quad \triangle A B D \cong \triangle A C D \quad \text { (SAS-criteria) } }

    {\color{Blue} \text{Hence}, \;\;\angle B=\angle C \quad ( \text{ Corresponding parts of Congruent triangles }) }

     

    Theorem 2 (Converse of Theorem 2): If two angles of a triangle are equal then the sides opposite to them are also equal..

    Given: Let a \triangle A B C\text{ in which }\angle B=\angle C

    To prove: A B=A C

    Construction: Draw AD, the bisector of \angle A, to meet BC at D.

    Proof: 

    {\color{Blue} \text{In } \triangle A B D\text{ and }\triangle A C D,\text{ we have,}}

    {\color{Blue} \angle B=\angle C \quad(\text { given })}

    {\color{Blue} \angle B A D=\angle C A D \quad(\text { by construction })}

    {\color{Blue} AD=AD \quad(\text { common })}

    {\color{Blue} \therefore \quad \triangle A B D \cong \triangle A C D \quad \text { (AAS-criteria) } }

    {\color{Blue} \text{Hence}, \;\;AB=AC \quad ( \text{ Corresponding parts of Congruent triangles }) }

    "Stay in the loop. Receive exam news, study resources, and expert advice!"

    Get Answer to all your questions