NCERT Class 9 Solutions: Triangles (Chapter 7) Exercise 7.3 – Part 3

Altitude of triangle

Give triangle of ABC, altibute is AB and base is BC

Give Altitude Triangle of ABC

Give triangle of ABC, altibute is AB and base is BC

In geometry, an altitude of a triangle is a line segment through a vertex and perpendicular to (i.e. forming a right angle with) a line containing the base (the opposite side of the triangle). This line containing the opposite side is called the extended base of the altitude.

Q-4 BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle ABC is isosceles

Solution:

Give a triangle ABC are two altitudes of BE and CF

Give a Triangle ABC

Give a triangle ABC are two altitudes of BE and CF

Given,BE and CF are two equal altitudes.

  • In ΔBEC and ΔCFB , BEC=CFB=90° (Altitudes) BC=CB (Common) BE=CF (Common)

  • Therefore, ΔBECΔCFB by RHS congruence condition.

  • Now, C=B (by Corresponding Parts of Congruent Triangles)Thus, AB=AC as sides opposite to the equal angles are equal.

Q-5 ABC is an isosceles triangle with AB=AC . Draw APBC to show that B=C.

Solution:

Give isosceles triangle ABC with AB=AC also AP ⊥ BC and ∠B = ∠C.

Give Isosceles Triangle ABC With AB=AC

Give isosceles triangle ABC with AB=AC also AP ⊥ BC and ∠B = ∠C.

  • Given, AB=AC In ΔABPandΔACP, APB=APC=90° (AP is altitude) AB=AC (Given) AP=AP (Common line)

  • Therefore, ΔABPΔACP by Right Angle-Hypostenuse-Side congruence condition.

  • Thus, B=C (by Corresponding Parts of Congruent Triangles)

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