Name: Proof of Basic Proportionality theorem
Uploaded: 2019-07-19
Duration: 3 min 10 s
Description: When a line is drawn parallel to one side, it intersects the other two sides of a triangle in the same ratio. Watch this video to learn more about them.

Question

If two isosceles triangle have a common base, prove that the line joining their vertices bisects the base at right angles.

Accepted Answer

$△ A B C$ and $△ D B C$ are two isosceles triangles with the common base that is $B C$ .

In $△ A B D$ and $△ A C D$ , we have,

$A B = A C$ [these are equal sides of the isosceles $△ A B C$ ];

$B D = C D$ [these are equal sides of the isosceles $△ D B C$ ];

$A D = A D$ [common side]

So, we have, $△ A B D ≅ △ A C D$ [ by SSS congruence rule]

_{$\Rightarrow ∠ B A E = ∠ C A E$ .} [CPCT]

Now, in $△ A B E$ and $△ A C E$ , we have,

$A B = A C$ [these are equal sides of the isosceles $△ A B C$ ];

$∠ B A E = ∠ C A E$ [proved above];

and, $A E = A E$ [common side].

So, we get, $△ A B E ≅ △ A C E$ $($ by SAS congruence rule $)$

Hence, we have $∠ B E A = ∠ C E A$ [CPCT]. ----(i)

Now, $∠ B E A + ∠ C E A = 180 °$ [They are linear pair of angles. Hence, their sum is $180 °$ ]

$\Rightarrow ∠ B E A + ∠ B E A = 180 °$ [By (i)]

$\Rightarrow 2 ∠ B E A = 180 °$

$\Rightarrow ∠ B E A = \frac{180 °}{2} = 90 °$ .

So, $∠ B E A = 90 ° = ∠ C E A$ .

Hence, $A D$ bisects the base $B C$ at right angles.

Therefore, if two isosceles triangle have a common base, then the line joining their vertices bisects the base at right angles.

If two isosceles triangle have a common base, prove that the line joining their vertices bisects the base at right angles.

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