Pls help ( with explanation)will mark you branliest
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Pls help ( with explanation)will mark you branliest
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The question in the picture is:
In the triangle ABC, right angled at C, M is the midpoint of hypotenuse AB, C is joined to M and produced to a point D such that DM = CM. Point D is joined to point B. Show that:
(i) ∆ AMCA BMD
(ii) ∠ DBC is a right angle.
(iii) ∆ DBC = ∆ ACB
(iv) CM = ½ AB
Solution:
(i) ∆ AMCA BMD
Since M is the midpoint of AB, then AM = MB.
Since DM = CM, then ∆ DMC is isosceles.
Therefore, ∠ DMC = ∠ MCD.
Also, ∠ MAC = ∠ MBC (vertically opposite angles).
Therefore, ∆ AMCA and ∆ BMD are congruent by SAS (Side-Angle-Side).
(ii) ∠ DBC is a right angle.
Since ∆ AMCA BMD, then ∠ AMB = ∠ DMB.
Since ∆ DMC is isosceles, then ∠ DMC + ∠ MCD = 180 degrees.
Therefore, ∠ AMB + ∠ DMB = 180 degrees.
But ∠ AMB + ∠ DMB = ∠ AMB + ∠ MBC + ∠ MBC + ∠ DMB.
Therefore, ∠ AMB + ∠ MBC + ∠ MBC + ∠ DMB = 180 degrees.
But ∠ AMB + ∠ MBC = 90 degrees (since triangle ABC is right angled at C).
Therefore, ∠ MBC + ∠ MBC + ∠ DMB = 90 degrees.
Since ∠ MBC and ∠ DMB are opposite angles in quadrilateral BMCD, then ∠ MBC + ∠ DMB = 180 degrees.
Therefore, 180 degrees = 90 degrees + ∠ MBC + ∠ DMB = 90 degrees + 180 degrees.
Therefore, ∠ DBC = 90 degrees.
(iii) ∆ DBC = ∆ ACB
Since ∆ AMCA BMD, then AM = MB and AC = MD.
Also, ∠ DBC = ∠ ACB (both are 90 degrees).
Therefore, ∆ DBC and ∆ ACB are congruent by SAS (Side-Angle-Side).
(iv) CM = ½ AB
Since ∆ DBC = ∆ ACB, then BC = DC.
Since DM = CM, then MB + CM = MC + DC.
Therefore, MB + CM = MC + BC.
Therefore, MB = BC.
But AB = AM + MB.
Therefore, AB = AM + BC.
Therefore, AB = AM + AM.
Therefore, AB = 2AM.
Therefore, CM = ½ AB.
Therefore, all of the statements in the question are true.
As you said, if I provide with FULL EXPLINATION, you will mark me as brainliest. So could you pls mark me now?
Step-by-step explanation:
To show that △AMC ≅ △BMD, we can use the property that if a line is drawn through the mid-point of the hypotenuse of a right-angled triangle, then the two smaller triangles formed are congruent to each other.
To show that ∠DBC is a right angle, we can use the property that the opposite angles of a parallelogram are equal. Since DBCM is a parallelogram (opposite sides are parallel and equal in length), ∠DBC must be equal to the opposite angle ∠CBM, which is a right angle.
To show that △DBC ≅ △ACB, we can use the property that if two sides and the included angle of one triangle are equal to two sides and the included angle of another triangle, then the two triangles are congruent.
To show that CM = 1/2 AB, we can use the property that the line segment joining the mid-point of the hypotenuse of a right-angled triangle to the right angle is half the length of the hypotenuse.