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For the third conclusion, the provided solution is:(double length center line) Extend CE to F such that EF = CE and connect AF as shown in the diagram below：It can be easily proven thatwhich implies ∠M = ∠ACE. Since AC = AM, triangle ACM is isosceles. Given that AE is the median to side CM, and combining this with the concurrency property of isosceles triangles, AE must be a height, meaning AE ⊥ CM (CE).However, why double the length of CE? Is there a simpler, more intuitive solution? Can we use this opportunity to cultivate students' geometric intuition and reasoning skills?The key to this problem is understanding a typical scenario from the converse of the isosceles triangle concurrency theorem, which states that "if the median and angle bisector to one side of a triangle coincide, then the triangle must be isosceles." In simpler terms, "two lines (the median and angle bisector) coinciding implies isosceles." In geometric notation:
In triangle ABC, if median CE bisects ∠ACB, prove that AC = BC.If students grasp this concept, they will find the solution to the third conclusion of the exam question intuitive! So, how can this problem be solved?Solution1：Of course, starting from the median and doubling its length (similar to the approach in the provided solution) is a clever method, which I won't elaborate on here.Solution2：Consider from the angle bisector CE, using the property of angle bisectors, we haveSolution3：Consider triangles AEC and BEC. AE = BE, CE is a common side, and ∠ACE = ∠BCE. This satisfies the SSA condition for triangle congruence. A common approach is to “cut long to make short.”Let’s assume we can cut CB to obtain CB' = CA. To prove AC = BC, we need to show B' coincides with B. If not, connecting EB', we observe △B'CE≌△ACE（SAS）, which leads to EB' = EA = EB. Connecting AB', we have:Then ∠AB'B = 90°, meaning BC ⊥ AB'. Since triangles B'CE and ACE are symmetric about CE, so there must be CE ⊥ AB', implying BC // CE, which contradicts the intersection point C. Hence, B' must coincide with B, proving AC = BC.Solution4：Again, using the "cut long to make short" method, draw EG ⊥ AC at G and EH ⊥ BC at H.From the property of the angle bisector, we know EG = EH. Given AE = BE, so Rt△AGE≌Rt△BHE（HL）, thus AG = BH. It can be easily shown that triangles CEG and CEH are also congruent, leading to CG = CH. Therefore, AG + CG = BH + CH, which implies AC = BC, completing the proof!Solution5: (Area formula) The area of triangle AEC = 1/2 × sin∠ACE × AC · EC, and the area of triangle BEC = 1/2 × sin∠BCE × BC · EC. Since the areas of triangles AEC and BEC are equal and ∠ACE = ∠BCE, it follows that AC = BC, completing the proof!Solution6: (Sine theorem) In triangles ACE and BCE, we apply the sine theorem, yielding AE/sin∠ACE = CE/sin∠A and BE/sin∠BCE = CE/sin∠B. Since ∠A + ∠B < 180°, it follows that ∠A = ∠B, hence AC = BC, completing the proof!Do you feel the beauty of geometric reasoning?![]()
