
In the adjoining figure,$AB = CD$, $CE = BF$ and $\angle ACE = \angle DBF$. Prove that $AE = DF$.
Answer
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Hint:
We are given that two sides and a pair of angles of the two triangles are equal and are asked to prove that the other two sides are also equal. Thus, we will first prove the two triangles to be congruent. Finally, we will prove the other two sides to be equal.
Complete step by step solution:
Here,
The two given triangles are $\Delta AEC$ and$\Delta DFB$.
Also,
$AB = CD$
$CE = BF$
$\angle ACE = \angle DBF$
Required to prove: $AE = DF$
Proof:
In $\Delta AEC$ and $\Delta DFB$
$AB = CD$
$\angle ACE = \angle DBF$
$CE = BF$
Thus, by the side angle side property of congruence, we get
$\Delta AEC \cong \Delta DFB$
Now,
If two sides of a triangle are equal and are congruent with each other.
Thus, clearly, we can say that the third pair of sides is also equal to each other. In other words, corresponding parts of congruent triangles (C.P.C.T.) are equal.
Thus,
$AE = DF$ (C.P.C.T.)
Hence, proved.
Additional Information:
Congruence of two triangles practically signifies that the two triangles are purely superimposable on each other.
Other than the property we used here, there are other properties which define and justify the congruence of the two triangles.
The other properties are:
1) Side-Side-Side (SSS): According to this property, if the three pairs of sides of the two triangles are equal, then we can say that the two triangles are congruent with each other.
2) Angle Side Angle (ASA): According to this property, if two pairs of angles and a pair of sides are equal, then we can say that the two triangles are congruent with each other.
Note:
We have only considered the pair of sides $AE, DF$ in accordance with this question. But other than that, the other pair of angles are also implied to be equal as soon as we evaluate the triangles to be congruent with each other.
We are given that two sides and a pair of angles of the two triangles are equal and are asked to prove that the other two sides are also equal. Thus, we will first prove the two triangles to be congruent. Finally, we will prove the other two sides to be equal.
Complete step by step solution:
Here,
The two given triangles are $\Delta AEC$ and$\Delta DFB$.
Also,
$AB = CD$
$CE = BF$
$\angle ACE = \angle DBF$
Required to prove: $AE = DF$
Proof:
In $\Delta AEC$ and $\Delta DFB$
$AB = CD$
$\angle ACE = \angle DBF$
$CE = BF$
Thus, by the side angle side property of congruence, we get
$\Delta AEC \cong \Delta DFB$
Now,
If two sides of a triangle are equal and are congruent with each other.
Thus, clearly, we can say that the third pair of sides is also equal to each other. In other words, corresponding parts of congruent triangles (C.P.C.T.) are equal.
Thus,
$AE = DF$ (C.P.C.T.)
Hence, proved.
Additional Information:
Congruence of two triangles practically signifies that the two triangles are purely superimposable on each other.
Other than the property we used here, there are other properties which define and justify the congruence of the two triangles.
The other properties are:
1) Side-Side-Side (SSS): According to this property, if the three pairs of sides of the two triangles are equal, then we can say that the two triangles are congruent with each other.
2) Angle Side Angle (ASA): According to this property, if two pairs of angles and a pair of sides are equal, then we can say that the two triangles are congruent with each other.
Note:
We have only considered the pair of sides $AE, DF$ in accordance with this question. But other than that, the other pair of angles are also implied to be equal as soon as we evaluate the triangles to be congruent with each other.
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