Advertisements
Advertisements
Question
ABC is an isosceles triangle with AB = AC. Drawn AP ⊥ BC to show that ∠B = ∠C.
Advertisements
Solution

Given: ABC is an isosceles triangle
In which AB = AC
To prove: ∠B = ∠C
Construction: Draw AP ⊥ BC.
Proof: In ∆ABP and ∆ACP,
∠APB = ∠APC ...(Each 90°) ...(By construction)
AB = AC ...(Given)
AP = AP ...(Common)
ΔABP ≅ ΔACP ...(By RHS congruence rule)
Hence, ∠B = ∠C ...(Corresponding parts of congruent triangles)
APPEARS IN
RELATED QUESTIONS
AD is an altitude of an isosceles triangles ABC in which AB = AC. Show that
- AD bisects BC
- AD bisects ∠A
BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle ABC is isosceles.
In two right triangles one side an acute angle of one are equal to the corresponding side and angle of the other. Prove that the triangles are congruent.
In the following figure, BA ⊥ AC, DE ⊥ DF such that BA = DE and BF = EC. Show that ∆ABC ≅ ∆DEF.

ABC is an isosceles triangle in which AC = BC. AD and BE are respectively two altitudes to sides BC and AC. Prove that AE = BD.
Prove that sum of any two sides of a triangle is greater than twice the median with respect to the third side.
In a right triangle, prove that the line-segment joining the mid-point of the hypotenuse to the opposite vertex is half the hypotenuse.
Two lines l and m intersect at the point O and P is a point on a line n passing through the point O such that P is equidistant from l and m. Prove that n is the bisector of the angle formed by l and m.
Line segment joining the mid-points M and N of parallel sides AB and DC, respectively of a trapezium ABCD is perpendicular to both the sides AB and DC. Prove that AD = BC.
ABC is a right triangle such that AB = AC and bisector of angle C intersects the side AB at D. Prove that AC + AD = BC.
