Problem-1: In the figure, ∆ABC is a triangle in which ∠ABC = 900, ∠BAC =480 and BD is perpendicular to AC. Find the remaining angles. Solution: Let remaining angles ∠ABD = x, ∠DBC = y and ∠BCD = z. Since BD⊥AC ∴ ∠ADB = ∠CDB = 900 Now, in ∆ABD ...

Read More »## If two angles of triangles are equal, then the sides opposite to the equal angles are equal.

If two angles of triangles are equal, then the sides opposite to the equal angles are equal. Particular enunciation: Let ABC be a triangle in which the ∠ACB = the ∠ABC. We have to prove that AB =AC. Proof: If AC and AB are equal, suppose that AB>AC. From BA ...

Read More »## In the figure, PM⊥QR, ∠QPM = ∠RPM and ∠QPR =900.

(a) Find the measure of ∠QPM. (b) What are the measure of ∠PQM and ∠PRM? (c) If PQ = 6 cm. Find the measure of PR. Solution: Given that, PM⊥QR, ∠QPM = ∠RPM and ∠QPR =900. (a) ∠QPM + ∠RPM = ∠QPR ⟹∠QPM + ∠QPM = ∠QPR [∵∠QPM = ...

Read More »## Prove that the angle opposite the greatest side of a triangle is also the greatest angle of that triangle.

Problem: Prove that the hypotenuse of a right angled triangle is the greatest side. Solution: General enunciation: We have to prove that the hypotenuse of a right angled triangle is the greatest side. Particular enunciation: Let ∆ABC be right angled triangle in which ∠ABC = right angle or 900 and ...

Read More »## ABC is an isosceles triangle and AB = AC. The side BC is extended up to D. Prove that AD>AB.

Solution: General enunciation: ABC is an isosceles triangle and AB = AC. The side BC is extended up to D. Prove that AD>AB. Particular enunciation: Given that, ABC is an isosceles triangle and AB = AC. The side BC is extended up to D. It is required to prove that ...

Read More »## Triangles| Geometry

Definition: The figure bounded by three line segments is a triangle .The line segments are known as the sides of the triangle. The point common to any two sides is known as vertex. The angle formed at the vertex is an angle of the triangle. The triangle has three sides ...

Read More »## If D is the middle point of the side BC of ∆ABC, prove that AB+AC>2AD.

General enunciation: If D is the middle point of the side BC of ∆ABC, prove that AB+AC>2AD. Particular enunciation: Suppose, in triangle ∆ABC, D is the mid-point of BC. Let us join (A, D). Let us prove that AB+AC>2AD. Construction: Let us expand AD up to DE such that, AD ...

Read More »## ABC is an isosceles triangle and AB=AC. The side BC is extend up to D. Prove that AD>AB.

General enunciation: ABC is an isosceles triangle and AB=AC. The side BC is extend up to D. Prove that AD>AB. Particular enunciation: Given that, ABC is an isosceles triangle and AB=AC. The side BC is extend up to D. It is required to prove that AD>AB. Proof: In ∆ ABC ...

Read More »## In the triangle ABC, AB>AC and the bisectors of the ∠B and ∠C intersect at the point P. Prove that PB>PC.

General enunciation: In the triangle ABC, AB>AC and the bisectors of the ∠B and ∠C intersect at the point P. Prove that PB>PC. Particular enunciation: In the triangle ABC, AB>AC and the bisectors of the ∠B and ∠C intersect at the point P. It is required prove that PB>PC. Proof: ...

Read More »## In the ∆ABC, the internal bisector of the ∠B and ∠C intersects at the point D. Prove that ∠BDC = 90°-½∠A.

Solution: General enunciation: In the ∆ABC, the internal bisector of the ∠B and ∠C intersects at the point D. We have to prove that ∠BDC = 90°-½∠A. Particular enunciation: In the ∆ABC, the internal bisector of the ∠B and ∠C is BD and CD intersects at the point D. ° ...

Read More »