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Diagonals AC and BD of quadrilateral ABCD intersect each other at P . Prove that ar (tri. APB)×ar (tri. CPD)=ar (tri. APD)×ar(tri. ABD)?
Verified Answer
Diagonals AC and BD of quadrilateral ABCD intersect each other at P . ...

Let ABCD is a quadrilateral having diagonal AC and BD. These diagonals intersect at point P.

Now area of a triangle  = 1/2 * base * height

=> Area(ΔAPB) * Area(ΔCPD)  = {1/2 * BP*AM}*{1/2 * PD*CN}

=> Area(ΔAPB) * Area(ΔCPD)  = 1/4 * BP*AM* PD*CN ....1

Again

      Area(ΔAPD) * Area(ΔBPC) = {1/2 * PD*AM}*{1/2 *CN*BP}

=> Area(ΔAPD) * Area(ΔBPC)  = 1/4 * BP*AM* PD*CN ....2

From equation 1 and 2, we get

Area(ΔAPB) * Area(ΔCPD)  = Area(ΔAPD) * Area(ΔBPC)
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Most Upvoted Answer
Diagonals AC and BD of quadrilateral ABCD intersect each other at P . ...
Proof:

Given:
- Quadrilateral ABCD with diagonals AC and BD intersecting at point P.

To prove:
- ar(tri. APB) × ar(tri. CPD) = ar(tri. APD) × ar(tri. ABD)

Proof:

Step 1: Introduction to the concept of area of triangles
- In geometry, the area of a triangle is given by the formula: Area = 1/2 * base * height.
- The base and height can be any two sides of the triangle, as long as the height is perpendicular to the base.

Step 2: Understanding the given quadrilateral and its diagonals
- We are given a quadrilateral ABCD with diagonals AC and BD intersecting at point P.
- Let's consider the triangles formed by these diagonals: triangle APB, triangle CPD, triangle APD, and triangle ABD.

Step 3: Explaining the key concept used in the proof
- The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
- In other words, if two triangles are similar, then the ratio of their areas is equal to the ratio of the squares of their corresponding sides.

Step 4: Applying the concept to the given triangles
- Let's consider the triangles APB and CPD.
- Since the diagonals AC and BD intersect at point P, we can conclude that triangles APB and CPD are similar (by Angle-Angle similarity criterion).
- Let's denote the sides of triangle APB as a, b, and c, and the sides of triangle CPD as d, e, and f.

Step 5: Using the concept to compare the areas of triangles APB and CPD
- Since triangles APB and CPD are similar, the ratio of their areas is equal to the ratio of the squares of their corresponding sides.
- Thus, (Area of APB) / (Area of CPD) = (a^2) / (d^2).

Step 6: Applying the concept to the other pair of triangles
- Let's consider the triangles APD and ABD.
- Since APD and ABD share the same base AD, the ratio of their areas is equal to the ratio of their corresponding heights.
- Let's denote the heights of triangle APD and ABD as h1 and h2, respectively.

Step 7: Comparing the areas of triangles APD and ABD
- Since the ratio of their areas is equal to the ratio of their heights, (Area of APD) / (Area of ABD) = h1 / h2.

Step 8: Bringing it all together
- We have established that (Area of APB) / (Area of CPD) = (a^2) / (d^2) and (Area of APD) / (Area of ABD) = h1 / h2.
- Rearranging the equations, we get (Area of APB) × (Area of CPD) = (a^2) × (h1^2) and (Area of APD) × (Area of ABD
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Diagonals AC and BD of quadrilateral ABCD intersect each other at P . Prove that ar (tri. APB)×ar (tri. CPD)=ar (tri. APD)×ar(tri. ABD)?
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Diagonals AC and BD of quadrilateral ABCD intersect each other at P . Prove that ar (tri. APB)×ar (tri. CPD)=ar (tri. APD)×ar(tri. ABD)? for Class 9 2024 is part of Class 9 preparation. The Question and answers have been prepared according to the Class 9 exam syllabus. Information about Diagonals AC and BD of quadrilateral ABCD intersect each other at P . Prove that ar (tri. APB)×ar (tri. CPD)=ar (tri. APD)×ar(tri. ABD)? covers all topics & solutions for Class 9 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Diagonals AC and BD of quadrilateral ABCD intersect each other at P . Prove that ar (tri. APB)×ar (tri. CPD)=ar (tri. APD)×ar(tri. ABD)?.
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