Showing posts with label Congruent Triangles. Show all posts
Showing posts with label Congruent Triangles. Show all posts

Sunday, December 27, 2015

Isosceles Trapeziums are Con-Cyclic

Problem : We need to prove that any isosceles trapezium is con-cyclic.






Solution :Here’s an isosceles trapezium with sides PQ and RS are parallel and PS = QR. A quadrilateral is cyclic if it’s opposite angles are supplementary (i.e. they add up to 180˚). So , we need to prove that QPS + QRS = 180˚ and PSR + PQR = 180˚.

Let us construct two perpendiculars, PU and QT, from point P and Q to segment RS.

Now, in ΔPSU and ΔQRT
PUS = QTR        by construction - both are right angles
PS = QR           Given trapezium is isosceles
PU = QT           perpendicular distance between two parallel lines
Thus, ΔPSU and ΔQRT are congruent by Right angle-Hypotenuse-Side (RHS) congruency rule
Now angles , PSU = QRT        Corresponding Parts of Congruent Triangles (CPCT)
Therefore ,angle PSR and angle QRS are equal. -------- (1)
Also, angle RQT is equal to angle SPU by CPCT. Adding right angles UPQ and TQP to the above angles, we get
RQT + TQP = SPU + UPQ
Thus, angle RQP is equal to SPQ -------- (2)
Adding equations 1 and 2 we get the following relations for angles
PSR + RQP = QRS + SPQ --------(3)
Since the sum of all the angles in a quadrilateral is 360˚,from equation (3)
PSR + RQP + QRS + SPQ = 360˚
2 (PSR + RQP) = 2 (QRS + QPS) = 360˚
PSR + RQP =QRS + QPS = 180˚
Since the opposite angles are supplementary, it can be concluded that an isosceles trapezium is a cyclic quadrilateral.

Friday, June 1, 2012

Squares on Sides of a Triangle

Squares ABDE and BCFG are drawn outside of triangle ABC : Prove that triangle ABC is isosceles if DG is parallel to AC.
This is a Java Applet created using GeoGebra from www.geogebra.org - it looks like you don't have Java installed, please go to www.java.com
Given that DG is parallel to AC. Draw a perpendicular from B to AC , this is also perpendicular to DG. Let the perpendicular intersect AC at P and DG at Q.

Since ∠ABP = 90° - ∠DBQ = ∠BDQ and AB = BD , the right triangles ABP and BDQ are congruent (by ASA Criteria) , hence AP = BQ (by CPCT).

Similarly , right triangles CBP and BGQ are congruent and BQ = PC. So , by the above , AP = CP and BP is perpendicular to AC , this implies that AB = BC , hence triangle ABC is isosceles.

Friday, March 2, 2012

Square and Circle - II

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Monday, September 19, 2011

Congruent Triangles - Right Angle Hypotenuse Side (RHS)




















This is a Java Applet created using GeoGebra from www.geogebra.org - it looks like you don't have Java installed, please go to www.java.com

Sunday, September 18, 2011

Congruent Triangles - Angle Side Angle (ASA)




















This is a Java Applet created using GeoGebra from www.geogebra.org - it looks like you don't have Java installed, please go to www.java.com

Congruent Triangles - Side Angle Side (SAS)




















This is a Java Applet created using GeoGebra from www.geogebra.org - it looks like you don't have Java installed, please go to www.java.com

Saturday, September 17, 2011

Congruent Triangles - Side Side Side (SSS)



















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