Answer
430.2k+ views
Hint: BM and CL are the two altitudes to AC and AB respectively. So, BM and CL altitudes are equal in length then BM=CL. Using this logic, we can find the type of the given triangle.
Complete step by step answer:
According to the question we have,
Given: In $\Delta $ABC, If BM=CL
In the given figure we have,
CL is perpendicular to AB and BM is perpendicular to AC
And,
$\angle M=90^\circ$ and $\angle L=90^\circ$
So, in the given figure, BM and CL are the two altitudes to AC and AB respectively
Since BM and CL altitudes are equal in length. Then,
$\Rightarrow BM=CL$
Therefore, the corresponding bases should also be of equal length which is intersecting at common vertex A.
$\Rightarrow AB=AC$
Hence the $\Delta $ABC must be isosceles $\Delta $. So, the correct answer is option B.
Note:
$BM=CL$, the corresponding bases should also be of equal length which is intersecting at common vertex A, then $AB=AC$. Therefore $\Delta $ABC must be an isosceles triangle. An isosceles triangle has one of the angles exactly $90^\circ$ and two sides which are equal to each other. Since the two sides are equal which makes the corresponding angle congruent. Thus, in an isosceles right triangle, two sides are congruent and the corresponding angle will be $45^\circ$. Each which sum of $90^\circ$. So, the sum of the three angles of the triangle will be $180^\circ$.
Complete step by step answer:
According to the question we have,
Given: In $\Delta $ABC, If BM=CL
In the given figure we have,
CL is perpendicular to AB and BM is perpendicular to AC
And,
$\angle M=90^\circ$ and $\angle L=90^\circ$
So, in the given figure, BM and CL are the two altitudes to AC and AB respectively
Since BM and CL altitudes are equal in length. Then,
$\Rightarrow BM=CL$
Therefore, the corresponding bases should also be of equal length which is intersecting at common vertex A.
$\Rightarrow AB=AC$
Hence the $\Delta $ABC must be isosceles $\Delta $. So, the correct answer is option B.
Note:
$BM=CL$, the corresponding bases should also be of equal length which is intersecting at common vertex A, then $AB=AC$. Therefore $\Delta $ABC must be an isosceles triangle. An isosceles triangle has one of the angles exactly $90^\circ$ and two sides which are equal to each other. Since the two sides are equal which makes the corresponding angle congruent. Thus, in an isosceles right triangle, two sides are congruent and the corresponding angle will be $45^\circ$. Each which sum of $90^\circ$. So, the sum of the three angles of the triangle will be $180^\circ$.
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