Question

Proove thales theorem​

Answer 1

If a line is drawn parallel to one side of a triangle intersecting the other two sides in distinct points, then the other two sides are divided in the same ratio.

TO PROVE:AP/PB = AQ/QC

PROOF:

CONSTRUCTION: Join the vertex B of ΔABC to Q and the vertex C to P to form the lines BQ and CP and then drop a perpendicular QN to the side AB and also draw PM⊥AC as shown in the given figure.

Now the area of ∆APQ = 1/2 × AP × QN (Since, area of a triangle= 1/2× Base × Height)

Similarly, area of ∆PBQ= 1/2 × PB × QN

area of ∆APQ = 1/2 × AQ × PM

Also,area of ∆QCP = 1/2 × QC × PM ………… (1)

Now, if we find the ratio of the area of triangles ∆APQand ∆PBQ, we have

ar(ΔAPC)/ar(ΔPBQ)=1/2×AP×QN/1/2×PB×QN=AP/QN

SIMILARLY ar(ΔAPQ)/ar(ΔQCP)=AQ/QC..........(2)

According to the property of triangles, the triangles drawn between the same parallel lines and on the same base have equal areas.

Therefore, we can say that ∆PBQ and QCP have the same area.

area of ∆PBQ = area of ∆QCP …………..(3)

Therefore, from the equations (1), (2) and (3) we can say that,

AP/PB = AQ/QC

HENCE PROVED

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Answer 2

$\huge\underbrace\mathtt\color{pink}{Hlw\:Mate}$

Basic Proportionality Theorem (BPT) or Thales theorem

$\Large\bold\blue{Statement}$

A straight line drawn parallel to a side of triangle intersecting the other two sides, divides the sides in the same ratio.

$\Large\bold{\underline{\pink{Proof}}}$

$\Large\bold{\underline{\green{Given: }}}$

In Δ ABC, D is a point on AB and E is a point on AC.

$\Large\bold\purple{To \:prove:}$

$\frac{AD}{DB} = \frac{AE}{EC}$

$\Large\bold{\underline{\orange{Construction:}}}$

Draw a line DE || BC

$\Large\bold{\underline{\blue{Steps:}}}$

∠ ABC = ∠ADE =∠1 [Corresponding angles are equal because DE || BC]

∠ACB = ∠AED = ∠2 [Corresponding Angel's are equal because DE || BC]

∠DAE = ∠BAE = ∠3 [Both triangles have a common angle]

ΔABC ~ ΔADE

$\frac{AB }{AD} = \frac{AC}{AE}$

$\frac{AD+DB}{AD} = \frac{AE+EC}{AE}$

$1 + \frac{DB}{AD} = 1 + \frac{EC}{AE}$

$\frac{DB}{AD} = \frac{EC}{AE}$

$\frac{AD}{DB} = \frac{AE}{EC}$

$\Large \green{\mid{\fbox{\tt Hence\:Verified }\mid}}$

$\Large\color{aqua}Happy\:Learning$

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