AD and BC are equal perpendiculars to a line segment AB (see Fig). Show that CD bisects AB.
AD and BC are equal perpendiculars to a line segment AB (see Fig). Show that CD bisects AB.
Solution and Explanation
Consider the triangles \( \triangle BOC \) and \( \triangle AOD \):
\[ \angle BOC = \angle AOD \quad \text{(Since they are vertically opposite angles)} \]
\[ \angle CBO = \angle DAO \quad \text{(Both are right angles, 90º)} \]
\[ BC = AD \quad \text{(Given in the problem)} \]
Using the AAS congruence criterion (Angle-Angle-Side), we conclude that:
\[ \triangle BOC \cong \triangle AOD \]
From CPCT (Corresponding Parts of Congruent Triangles), we get:
\[ BO = AO \]
Therefore, the line \( CD \) divides the segment \( AB \) into two equal parts.
Top Questions on Congruence of Triangles
AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that
(i) AD bisects BC
(ii) AD bisects ∠A.
- CBSE Class IX
- Mathematics
- Congruence of Triangles
∆ ABC and ∆ DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see Fig.). If AD is extended to intersect BC at P, show that
(i) ∆ABD ≅ ∆ACD
(ii) ∆ABP≅ ∆ACP
(iii) AP bisects ∠A as well as ∠D.
(iv) AP is the perpendicular bisector of BC.
- CBSE Class IX
- Mathematics
- Congruence of Triangles
In right triangle ABC, right angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that DM = CM. Point D is joined to point B (see Fig. 7.23). Show that:
(i) ∆ AMC ≅ ∆ BMD
(ii) ∠ DBC is a right angle.
(iii) ∆ DBC ≅ ∆ ACB (iv) CM = \(\frac{1}{2}\) AB
- CBSE Class IX
- Mathematics
- Congruence of Triangles
AB is a line segment and P is its mid-point. D and E are points on the same side of AB such that ∠ BAD = ∠ ABE and ∠ EPA = ∠ DPB (see Fig). Show that
(i) ∆ DAP ≅ ∆ EBP
(ii) AD = BE
- CBSE Class IX
- Mathematics
- Congruence of Triangles
- Want to practice more? Try solving extra ecology questions todayView All Questions