\(\triangle ABC\) and \(\triangle DBC\) are two isosceles triangles on the same base \(BC\) and vertices \(A\) and \(D\) are on the same side of \(BC\). If \(AD\) is extended to intersect \(BC\) at \(P\), show that:
- \(\;\triangle ABD \cong \triangle ACD\)
- \(\;\triangle ABP \cong \triangle ACP\)
- \(\;AP\) bisects \(\angle A\) as well as \(\angle D\)
- \(\;AP\) is the perpendicular bisector of \(BC\)
📘 Concept & Theory Concept Used ›
- Properties of an isosceles triangle
- SSS Congruence Rule
- SAS Congruence Rule
- CPCT (Corresponding Parts of Congruent Triangles)
- Linear Pair Property
- Perpendicular Bisector Property
📝 theory insight ›
When two triangles are isosceles on the same base, symmetry plays an important role. The line joining the vertices of the two isosceles triangles often becomes:
- An angle bisector
- A perpendicular bisector
- A common axis of symmetry
This problem combines multiple geometric ideas together and is very important for proving properties related to congruence and symmetry in triangles.
🗺️ Solution Roadmap Step-by-step Plan ›
First prove \(\triangle ABD \cong \triangle ACD\) using SSS rule.
Use CPCT to obtain equal angles at \(A\) and \(D\).
Then prove \(\triangle ABP \cong \triangle ACP\) using SAS rule.
Again apply CPCT to obtain equal angles at \(P\).
Use linear pair property to show both angles are \(90^\circ\).
Conclude that \(AP\) is perpendicular to \(BC\) and bisects \(BC\).
📊 Graph / Figure Graph / Figure ›
✏️ Solution Part (i) To Prove: \(\small \triangle ABD \cong \triangle ACD \) ›
- Given \[\small \begin{aligned} AB &= AC \quad \text{(Since } \triangle ABC \text{ is isosceles)}\\ DB &= DC \quad \text{(Since } \triangle DBC \text{ is isosceles)} \end{aligned} \]
- To Prove \[\small \triangle ABD \cong \triangle ACD \]
- Proof
- In triangles \(\small \triangle ABD\) and \(\small \triangle ACD\):
- \[\small \begin{aligned} AB &= AC \quad \text{(Given)}\\ BD &= CD \quad \text{(Given)}\\ AD &= AD \quad \text{(Common side)} \end{aligned} \]
- Since all the three corresponding sides are equal,
- by SSS Congruence Rule.\[\small \triangle ABD \cong \triangle ACD\]
Using CPCT:
- \[\small \angle BAD = \angle DAC \] and \[\small \angle BDA = \angle ADC \]
- Since \(\small A,D,P\) are collinear,\[\small \angle BAP = \angle PAC\]
- Also,\[\small \angle BDP = \angle PDC\]
✏️ Solution Part (ii) To Prove \(\small \triangle ABP \cong \triangle ACP\) ›
To Prove:
- \[\small \triangle ABP \cong \triangle ACP\]
Proof:
- In triangles \(\small \triangle ABP\) and \(\small \triangle ACP\):
- \[\small \begin{aligned} AB &= AC \quad \text{(Given)}\\ AP &= AP \quad \text{(Common side)}\\ \angle BAP &= \angle PAC \quad \text{(From CPCT)} \end{aligned} \]
- Therefore, by SAS Congruence Rule \[\small \triangle ABP \cong \triangle ACP\]
✏️ Solution Part (iii) To Prove: \(AP\) bisects \(\angle A\) and \(\angle D\) ›
\(\small AP\) bisects \(\small \angle A\) and \(\small \angle D\).
- From CPCT obtained earlier:\[\small \angle BAP = \angle PAC\]
- Hence, \(\small AP\) bisects \(\small \angle A\).
- Also, \[\small \angle BDP = \angle PDC\]
- Therefore, \(AP\) also bisects \(\angle D\).
✏️ Solution Part (iv) To Prove: \(\small AP\) is the perpendicular bisector of \(\small BC\). ›
To Prove:
- \(AP\) is the perpendicular bisector of \(BC\).
- From \[\small \triangle ABP \cong \triangle ACP\]
- by CPCT: \[\small \angle APB = \angle APC\]
- Since \(\small B,P,C\) are collinear, \[\small \angle APB + \angle APC = 180^\circ\]
- But, \[\small \angle APB = \angle APC\]
- Therefore,
- \[\small \begin{aligned} 2\angle APB &= 180^\circ\ \angle APB &= 90^\circ \end{aligned} \]
- Hence, \[\small AP \perp BC\]
- Again from CPCT:\[\small BP = PC\]
- Thus, \(\small P\) is the midpoint of \(\small BC\).
- Since \(\small AP\) is perpendicular to \(\small BC\) and passes through its midpoint,
- \[\small AP \text{ is the perpendicular bisector of } BC\]
🎯 Exam Significance Exam Significance ›
This problem is highly important for:
- CBSE Board Examinations
- School-level Olympiads
- NTSE Foundation Preparation
- JEE Foundation Geometry
- Logical proof-writing practice
Students learn how one congruence proof can be repeatedly used to derive multiple geometric conclusions such as:
- Angle bisector
- Perpendicularity
- Midpoint relation
- Symmetry in triangles
Such multi-concept questions are frequently asked in board examinations because they test theorem application, reasoning ability, and systematic proof-writing.